Question

If local maximum of $f(x) = \frac{ax+b}{(x-1)(x-4)}$ exists at $(2,-1)$, then $a+b=$

• A. 0
• B. -1
• C. 1
• D. 2

Hint:

For a local extremum at $(x_0,y_0)$: $f(x_0)=y_0$ and $f'(x_0)=0$. Use these to form equations and solve for unknowns.

Answer 1

The Correct Option is C

Step 1: Use the point condition.
For a local maximum at $(2,-1)$: \[ f(2) = \frac{2a+b}{(2-1)(2-4)} = \frac{2a+b}{1\cdot (-2)} = -1 \implies 2a+b=2. \]
Step 2: Use derivative condition.
The first derivative: \[ f(x) = \frac{ax+b}{x^2-5x+4} \implies f'(x) = \frac{a(x^2-5x+4) - (ax+b)(2x-5)}{(x^2-5x+4)^2}. \] Set numerator zero at $x=2$: \[ a(4-10+4) - (2a+b)(4-5) = -2a -(-1)(2a+b)=0 \implies b=0. \]
Step 3: Find $a+b$.
From $2a+b=2$, $b=0 \implies a=1$.
Thus, \[ a+b = 1+0 = 1. \]